All categories

2 years ago

Plz help asasp it would be appreciated

Mathematics

2 answers:

Reil [10]2 years ago

7 0

Answer:

6x+10y=1060

6= the cost of tickets bought before the dance

x=the number of tickets bought before the dance

10= the cost of tickets bought at the doors

y= the number of tickets bought at the door

1060= total amount of money made from selling tickets

Sidana [21]2 years ago

4 0

This will be your answer

You might be interested in

on a average saturday, 15/100 of tickets sold are child tickits. if there were 460tickits sold saturday, how many were sold to c

liraira [26]

The answer is 69, because 15% of 460 is 69

Hope this helps!

6 0

3 years ago

Using the method described in part F, help Kevin change 9/2= x/24 to an equation without fractions. Show your work. Don't solve

soldi70 [24.7K]

Answer:

108 = x

Step-by-step explanation:

multiple both sides by 24 and simplify

9 x 24 = 216/2=x

108 = x

3 0

3 years ago

Read 2 more answers

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

Write 9-³ in standard form

ValentinkaMS [17]

729 wish you well on your assignment ~JyMarkus

5 0

2 years ago

Find the general solution of {y}''-3y=8e^{3t}+4sin(t) .

babunello [35]

Answer:

$y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint$

Step-by-step explanation:

We are given that linear differential equation

$y''-3y=8e^{3t}+4 sint$

Auxillary equation

$D^2-3=0$

$D=\pm \sqrt3$

C.F= $C_1e^{\sqrt3t}+C_2e^{-\sqrt3}$

P.I= $\frac{8e^{3t}+4sin t}{D^2-3}$

P.I= $\frac{8e^{3t}}{9-3}+4\frac{sint }{-1-3}$

P.I= $e^{ax}{\phi (D+a)}$ and $P.I=\frac{sinax}{(\phi D)}$ where D square is replace by - a square

P.I= $-\frac{4}{3}e^{3t}- sint$

Hence, the general solution

G.S=C.F+P.I

$y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint$

3 0

3 years ago